MHB What is the probability of selecting two males given that both are grey?

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The problem involves calculating the probability of selecting two male mice given that both are grey, with a population of 6 grey females, 2 grey males, 6 white females, and 2 white males. The correct approach involves using conditional probability, focusing only on the grey mice, which total 8. The probability of the first grey mouse being male is 2 out of 8, or 1/4, and if the first is male, the probability that the second is also male is 1 out of 7. Therefore, the overall probability of selecting two grey males is calculated as (1/4) * (1/7), resulting in a final probability of 1/28. This method effectively narrows the sample space to only the relevant grey mice.
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So I am having more problems with my homework...

For this problem, assume there are 6 grey females, 2 grey males, 6 white females, and 2 white males. Two mice are randomly selected. What is the probability of selecting two males given that both are grey?

So for selecting two males, I was thinking of 4/12, since there are 4 males in total out of 12. Then for the grey I was thinking of 8/12, since there are 8 grey mice. However, that did not work once I plugged it into the formula...

I thought of using combinations, but I am not sure how that would work and how I should write that. I was thinking of C(4,2)/C(12/2) for the intersection of grey mice and male, and for the "given that both are grey" C(8,2)/C(12,2)...?

Tbh, I am not sure of what I am doing...
 
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So, conditional probability has the effect of "narrowing your world", so to speak. The two mice are both grey, so you can completely ignore the whites.

You pick the first grey mouse without replacement (I think that's understood, though it's good to be clear, because it does change probabilities). What's the probability that it's male? Now pick the second grey mouse, assuming that the first one you picked was male. What's the probability that this second one is male?
 
Given that both mice are gray then we can ignore the other mice. There are 6 gray females and 2 gray males for a total of 8 gray mice. The probability the first mouse drawn is male is 2/8= 1/4. There are then 6 gray females and 1 gray male. The probability the second mouse drawn is male is 1/7. Given that the two mice drawn are gray the probability they are male is (1/4)(1/7)= 1/28.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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